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pishuonlain [190]
3 years ago
5

A small family home in Tucson, Arizona has a rooftop area of 2667 square feet, and it is possible to capture rain falling on abo

ut 61.0% of the roof. A typical annual rainfall is about 14.0 inches. If the family wanted to install a tank to capture the rain for an entire year, without using any of it, what would be the required volume of the tank in m3 and in gallons? How much would the water in a full tank of that size weigh (in N and in lbf)?
Engineering
1 answer:
Kazeer [188]3 years ago
7 0

Answer:

volume  = 53.747 m3 = 14198.138 gal

weight = 526652 N = 118396.08 lbf

Explanation:

We know that volume of water

volume  =  A'\times H

where A' = 61% of A

              = 0.61\times 2667 = 1626.87 sq ft

volume  =  1626.87 \times (\frac{14}{12} ft)

               =1898.015 ft^3

in\ m^3 = \frac{ 1898.015}{35.315} =   53.7457 m^3

in\ gallon = 1898.015 \times 7.481 = 14198.138 gallon

weight = \rho Vg

       = 1000\times 53.74\times 9.8

             =526652 N

In\ lbf =  \frac{526652}{4.448} = 118396.08 lbf

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N = mg + F₃"

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N = 343 N + 8.82 N

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F = 155.41 N - 0.4 × 351.82 N

F = 155.41 N - 140.728 N

F = 14.682 N

Since F = ma, where a = acceleration of block,

a = F/m = 14.682 N/35 kg = 0.42 m/s²

To find the distance the block moved, x we use the equation

x = ut + 1/2at² where u = initial speed of block = 0 m/s, t = time = 10 s and a = acceleration of block = 0.42 m/s²

Substituting the values of the variables into the equation, we have

x = ut + 1/2at²

x = 0 m/s × 10 s + 1/2 × 0.42 m/s² × (10 s)²

x = 0 m + 1/2 × 0.42 m/s² × 100 s²

x = 0.21 m/s² × 100 s²

x = 21 m

So, the distance moved by the block is 21 m.

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